1)

Let [ε0] denote the dimensional formula of the permittivity of vacuum. If M= mass, L= length , T= time and A= electric current then,

  


A) [ϵ0]=[M1L3T2A]

B) [ϵ0]=[M1L3T4A2]

C) [ϵ0]=[M2L2T1A2]

D) [ϵ0]=[M1L2T1A2]

Answer:

Option B

Explanation:

from COULOMB'S law  

                           F=14πϵ0q1q2R2

ϵ0=q1q24πFR2

substituting the units we have ,

 ϵ0=C2N.m2=[AT]2[MLT2][L2]=

                                         [M1L3T4A2]